If x, y, and k are positive and x is less than y, then (x + k)/(y + k)

Question

If x, y, and k are positive and x is less than y, then  \frac{x + k}{y + k}  is

A. 1
B. greater than x/y
C. equal to x/y
D. less than x/y
E. less than x/y or greater than x/y, depending on the value of k

PS87502.01
Quantitative Review 2020 NEW QUESTION

BrentGMATPrepNow · Accepted Answer

Key concept (also covered in video below):
 https://i.imgur.com/6q0g2tM.png

Since x < y, we know that  x/y is less than 1

So,  \frac{x + k}{y + k}  will be closer to 1 than x/y is.

In other words,  \frac{x + k}{y + k}  must be greater than x/y

Answer: B

Cheers,
Brent

IanStewart · Answer

These comments are not correct, so I'll clarify. First, a "fraction" is just a way to write a number, by using a numerator and a denominator (as opposed to, say, using decimal notation). When we call a number a "fraction", we're not saying anything about how big the number is -- "18/5" is just as much a fraction as "1/2" is. You both mean to describe a fraction with an overall value specifically between 0 and 1. Then if the numerator and denominator are both positive, when we add a positive number k to both the numerator and denominator, the overall value of the fraction will increase. So if we start with, say, the fraction 2/3, then if k > 0, it is always true that 2/3 < (2 + k)/(3 + k) < 1 But if you start with a fraction bigger than 1, with a positive numerator and denominator, the opposite of the above is true -- adding a positive number to the numerator and denominator will decrease the fraction's value. 3/2 is larger than 6/5, for example. Second, it's not technically correct to say that this principle applies to "positive fractions" (even with the restriction that the "fraction" has a value between 0 and 1). What matters is not whether the fraction itself is positive. What matters is that the numerator and denominator are both individually positive. The fraction (-1)/(-5) is a "positive fraction", since it is equal to 1/5, but if you add 2 to the numerator and denominator of (-1)/(-5), you get 1/(-3), which is now smaller than what we started with: its value does not increase. The general rule is this: if x, y, a and b are all positive, then one of the following three things is always true: x/y < (x + a)/(y + b) < a/b or x/y > (x + a)/(y + b) > a/b or x/y = (x + a)/(y + b) = a/b so in other words, when everything is positive, if we add a to the numerator and b to the denominator of a fraction, the fraction's overall value moves towards a/b. That's a principle everyone learns, usually in a very different way, when studying weighted averages (at least from a source that explains weighted averages well). But if there might be negative values anywhere, none of the above applies.

Archit3110 · Answer

let x=5 ,y=6 and k =1
solve
 \frac{x + k}{y + k}  = 6/7
and x/y=5/6
we see that 6/7>5/6 
IMO B

shaonkarim · Answer

x<y
So, x/y= less than 1
If x=1
 y=2
x/y=1/2=.5

Now, if we increase both numerator and denominator by same number then result must be increased.
1+1/2+1=2/3=.67

Answer is B

Posted from my mobile device

ScottTargetTestPrep · Answer

Let's compare (x + k)/(y + k) and x/y:

Is (x + k)/(y + k) > x/y?

Is (x + k)y > (y + k)x?

Is xy + ky > xy + kx?

Is ky > kx?

Is y > x? Yes!

Since y > x, all the inequalities above hold. So, (x + k)/(y + k) is greater than x/y.

Notice that we multiplied each side of the inequalities by constants several times, but all of x, y and k are positive; so we never had to worry about reversing the direction of the inequality.

CrackverbalGMAT · Answer

One way to solve this question is to know that if you add the same constant to the numerator and denominator of a positive fraction, then the value of the fraction increases and will always tend to 1. But even if we did not know this, we can easily solve the question by using the answer choices. Remember this is a PS question, so there have to be only ONE possible answer.

We have x, y and k to be positive and x < y. The question asks us for (x + k)/(y + k)

A. If (x + k)/(y + k) = 1, then x = y. This is not possible since x < y
B. If (x + k)/(y + k) > x/y, cross multiplying (since x and y are positive), we get xy + yk > xy + xk. Cancelling xy on both sides, we get y > x or x < y. So B works.

Hope this helps!

Aditya
CrackVerbal Quant Expert

MHIKER · Answer

let x= 2, y= 4, and y = 1
Then 2/4 = 0.5
(2+1)/(4+1) = 0.6

so the value is greater than x/y.

The answer is B.

sujoykrdatta · Answer

Obviously there is no reason why the answer would be equal to 1 - so let us ignore that. We basically need to compare the new fraction (x+k)/(y+k) with x/y Method 1 Given x, y and k are positive and x < y => x/y is a fraction less than 1 For example: Let us look at 2/3 (= 0.67) If we add 1 to both numerator and denominator, we have 3/4 (= 0.75) Thus, the new fraction is greater than the initial one. The reason for this is as follows: Since x < y, when we add the same constant k to x and to y, the percent increase in the value of x is more than that for y Thus, the numerator increases more in comparison to the denominator. Thus, the new fraction increases in value Thus, if x, y, k are positive and x < y, we have: (x+k)/(y+k) > x/y Method 2 We need to compare (x+k)/(y+k) and x/y The symbol between them would be '>' or '<' or '=' Let the symbol be # Thus: (x+k)/(y+k) # x/y Since all terms are positive, we can easily cross-multiply: => xy + yk # xy + xk => yk # xk Cancelling k from both sides (since k is positive): y # x But, we know that y > x Thus: '#' represents '>' Thus: (x+k)/(y+k) > x/y Answer B

Kinshook · Answer

Asked: If x, y, and k are positive and x is less than y, then  \frac{x + k}{y + k}  is

x<y
x/y < 1
For example
x = 2; y=3; k = 1
x =2 < y=3
x/y = 2/3
(x+k)/(y+k) = 3/4 > 2/3 = x/y

IMO B

dc2880 · Answer

Can the method in which you used simplification and # and to represent =, >, or < be used in any situation to find the relationship between two functions/equations?

shrutisanghi · Answer

Assume x/y as 2/3. this we know is equal to 0.67.
now take k as 7. (2+7)/(3+7)=9/10. This is equal to 0.9.

Hence, x/y<(x+k)/(y+k).

Yush12 · Answer

Using concepts I learnt from TTP,

If x is less than y, then x/y is a positive proper fraction.

Positive Proper fractions have a value between 0 and 1.

K, a positive integer, when added to the numerator and the denominator of a fraction brings the value of a fraction closer to 1.

In this case, as x/y is a fraction that has a value between 0 and 1, (x+k)/(y+k) will be closer to 1 and greater than x/y.

Therefore, (x/y) <

Ans. B

avigutman · Answer

We are  adding  the same thing to both the numerator and denominator, so the ratio of those additive quantities is 1:1. Therefore, the starting ratio will move along the number line toward 1 (since 1:1 is 1). 
Since the starting ratio (x/y) was between 0 and 1 (x and y are positive, x is less than y), moving toward 1 means it is increasing.

beeblebrox · Answer

BrentGMATPrepNow  /  chetan2u , sir can you elaborate further on the properties of fractions? For instance what happens when a constant is subtracted/added from/to numerator(N) and denominator(d) when N<D & N>D etc.?

BrentGMATPrepNow · Answer

Happy to help. Please note that the following only applies to positive fractions. If we have a fraction in the form N/D, where N > D, then N/D > 1 So, (N - k)/(D - k) will be greater than N/D (for positive values of k) For example, (10 - 4)/(7 - 4) > 10/7 Conversely, if we have a fraction in the form N/D, where N < D, then N/D < 1 So, (N - k)/(D - k) will be less than than N/D (for positive values of k) For example, (3 - 2)/(7 - 2) < 3/7

woohoo921 · Answer

ScottTargetTestPrep Thank you for this helpful explanation. I have two questions: Question 1: Why do you put y(x+k) on the left-hand side of > vs x(y+k)... is this the order of operations for cross-multiplying fractions in inequalities? Question 2: If you had x>y as an end result, what would be the next step when looking at the answer choices? You just know that (x+k)/(y+k) is not greater than x/y but is it equal to 1 (choice A), equal to x/y (choice C) etc. Would you repeat the process setting (x+k)/(y+k)=1 and if that doesn't work (x+k)/(y+k)=x/y I just reasoned through the question per Avi Gutman's approach knowing that adding 1 brings me closer to one but not quite at 1. However, I was curious about the algebra approach here. Thank you

egmat · Answer

Hey woohoo921 Let me try to elaborate the points from Scott’s crisp solution. I will answer both your questions one by one. 😊 Your First Question: “Why do you put y(x+k) on the left-hand side of > vs x(y+k)” This is due to the principle of cross-multiplication. Let’s take a closer look at the process of cross-multiplication in the context of inequalities. It is actually composed of two different operations that just happen at the same time. Let me try to explain the same to you through a simple example in which I will show you both these operations separately first and then together. Example: Consider ¾ > ½. ----------(I) We will try to remove the denominators from both fractions one by one. Step 1: Let’s start with the denominator 2 from ½. Multiplying inequality-I above throughout by 2, we get: ¾ × 2 > ½ × 2 ¾ × 2 > 1 --------(II) Note: Since 2 is a positive number, multiplying both sides of the inequality by 2 will not affect the sign of inequality. Observe how the denominator from the right-hand side (2 of 1/2) went to get multiplied with ¾ on the left-hand side. The numerator 3 did not change its place. Step 2: Next, let’s try to remove the denominator 4 from ¾. Multiplying inequality-II above throughout by 4, we get: ¾ × 2 × 4 > 1 × 4 3 × 2 > 1 × 4 6 > 4 Note: Since 4 is a positive number, multiplying it to both sides of the inequality does not impact the sign of the inequality. Observe how the denominator from the left-hand side (4 of ¾) went to get multiplied with 1 on the right-hand side. The numerator 1 did not change its place. SUMMARY: In both these steps, we took the positive denominators from each side of the inequality and multiplied them to the numerator on the other side without changing the inequality sign. The product in each case was written on the same side as the numerator. Everything we did in two steps above could also have been done in a single step. (This is what we call cross-multiplication). We could straight go from ¾ > ½ to 3 x 2 > 1 x 4. This is precisely what Scott did. He went from (x + k)/(y + k) > x/y to y(x + k) > x(y + k). Your Second Question: “What if we came to “Is x > y” at the end?” Let’s try to create this scenario. Let’s suppose Scott had started with “Is (x + k)/(y + k) < x/y?” Then, we would have: Is (x + k)/(y + k) < x/y? Is (x + k)y < (y + k)x? Is xy + ky < xy + kx? Is ky < kx? Is y < x? Now, we are given that y > x. Thus, the answer to the last question (Is y < x?) is NO! And since the last inequality does not hold true, all the above inequalities would not hold true as well. So, (x + k)/(y + k) is not less than x/y. Now, since < is rejected, we have two possibilities left: > or =. But how do we decide which one? Well, for now, let’s just put a “?” between the two fractions being compared. This “?” will eventually be either “>” or “=”. Here we start: (x + k)/(y + k) ? x/y Since, all terms are positive, we can cross-multiply without affecting the sign of the inequality. Let’s do that. y(x + k) ? x(y + k) xy + ky ? xy + kx ky ? kx y ? x But we already know that y > x! So “?” must actually represent “>”. Thus, (x + k)/(y + k) > x/y. And we’re done! Takeaways: Cross-multiplying fractions in an inequality involves keeping each numerator as is, and bringing the denominators from the other side for multiplication. The inequality sign remains the same if the denominator being moved is positive. For example, 1/3 > ¼. When we cross-multiply, the sign remains the same and we get 1 x 4 > 1 x 3. For questions like this, where you need to compare two entities, the easiest approach is to start with a “?” between the two entities. As you would simplify this (using normal rules of equations and inequalities), you would be able to use some given information to infer what “?” stands for out of >, <, or =. Hope this helps! Best Regards, Ashish Quant Expert, e-GMAT

totaltestprepNick · Answer

B https://www.youtube.com/watch?v=IDfjDPCsl84 Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience tutoring@totaltestprep.net

If x, y, and k are positive and x is less than y, then (x + k)/(y + k)

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